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Okay, here is my hypothetical situation. Lets say there is a cylinder

  1. Apr 17, 2008 #1
    Okay, here is my hypothetical situation. Lets say there is a cylinder and we're looking at it with the round side facing us. Its diameter is 48in. There is a 48in rod going through that cylinder(with the ends shaped so that it conforms to the circumference of the cylinder) from the left side to the right side perfectly in parallel with the floor, which is level. Now imagine that rod can slide to the left and right 1 inch. Still following? OK, next roll that cylinder clockwise until the rod, which is now extended its full 1 inch to the right, touches the floor. Now imagine a second rod going through that cylinder exactly the same as the first, parallel to the floor and extended to the right 1 inch. would it have the required rotational force to roll the cylinder the rest of the way clockwise until the rod was vertical?

    My thought is the only force required is enough to take the rod touching the floor, and tilt it to the right the few degrees required with the fulcrum being the point where it is touching the floor. Am I right that the point touching the floor would remain pretty much stationary while the rod tipped?

    I wish I had one more physics class, this doesn't seem like that hard of a question to me, but I just dont know the formulas required.
  2. jcsd
  3. Apr 18, 2008 #2
    Ignore the cylinder. Say you made a X out of the two rods, one is the near vertical one on the floor, the other is the horizontal one offset to the right, with the same constraints of 1 inch of movement. At some near veritical angle, the X will fall to the right instead of the left. So you just need to calculate the angle the near vertical rod is at, and then the torque forces on the X to see which way it would fall.
  4. Apr 30, 2008 #3
    lol now that you put it that way, it's kinda easy. thanks.
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